document.write( "Question 1142720: A spherical ballon is being filled with air at a constant rate of 2cm^3 per second. By using the chain rule dr/dt equals to 1/2(pi)(r^2) where t is in seconds. The rate at which the radius is increasing when r=54 is 1/5832pi. The balloon will burst when the radius reaches 100cm. Find the rate at which the surface area is increasing at that point in time. \n" ); document.write( "
Algebra.Com's Answer #763449 by Alan3354(69443)![]() ![]() You can put this solution on YOUR website! A spherical ballon is being filled with air at a constant rate of 2cm^3 per second. By using the chain rule dr/dt equals to 1/2(pi)(r^2) where t is in seconds. The rate at which the radius is increasing when r=54 is 1/5832pi. The balloon will burst when the radius reaches 100cm. Find the rate at which the surface area is increasing at that point in time. \n" ); document.write( "------------------- \n" ); document.write( " \n" ); document.write( "dV/dt = 4pi*r^2*dr/dt = 2 \n" ); document.write( "dr/dt = 1/(2pi*r^2) which is 1/(5832pi) at r = 54 \n" ); document.write( "---------- \n" ); document.write( "dr/dt at r=100 is 1/(2pi*10000) = 1/(20000pi) \n" ); document.write( "================= \n" ); document.write( "SA = 4pi*r^2 \n" ); document.write( "dSA/dt = 8pi*r*(dr/dt) \n" ); document.write( "dSA/dt = 8pi*100/20000pi \n" ); document.write( "= 1/25 sq cm/sec \n" ); document.write( "================ \n" ); document.write( "\"...at that point in time.\" is superfluous. \n" ); document.write( "\"at that point.\" or \"at that time.\" is better and sufficient. \n" ); document.write( "--- \n" ); document.write( "No need for \"at the point in space, or in the space-time continuum\" either.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |